preliminaries of real analysis

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Chapter A

Preliminaries of Real Analysis

A principal objective of this largely rudimentary chapter is to introduce the basic set-theoretical nomenclature that we adopt throughout the text. We start with an intuitive discussion of the notion of set, and then introduce the basic operations on sets, Cartesian products, and binary relations. After a quick excursion to order theory (in which the only relatively advanced topic thatwe cover is the completionof apartial order), functions are introduced as special cases of binary relations and sequences as special cases of functions. Our coverage of abstract set theory concludes with a brief discussion of the Axiom of Choice and the proof of Szpilrajns Theorem on the completion of a partial order.

Weassumehere that the reader is familiarwith the elementary properties of the real numbers and thus provide only a heuristic discussion of the basic number systems. No construction for the integers is given, in particular. After a short elaboration on ordered elds and the Completeness Axiom, we note without proof that the rational numbers form an ordered eld and the real numbers form a complete ordered eld. The related discussion is intended to be read more quickly than anywhere else in the text.

We next turn to real sequences. These we discuss relatively thoroughly because of the important role they play in real analysis. In particular, even though our coverage will serve only as a review for most readers, we study here the monotonic sequences and subsequential limits with some care, and prove a few useful results, such as the Bolzano-Weierstrass Theorem and Dirichlets Rearrangement Theorem. These results will be used freely in the remainder of the book.

The nal section of the chapter is nothing more than a swift refresher on the analysis of real functions. First we recall some basic denitions, and then, very quickly, we go over the concepts of limits and continuity of real functions dened on the real line. We then review the elementary theory of differentiation for single-variable functions, mostly through exercises. The primer we present on Riemann integration is a bit more leisurely.

4 | Chapter A Preliminaries

In particular, we give a complete proof of the Fundamental Theorem of Calculus, which is used in the remainder of the book freely. We invoke our calculus review also to outline a basic analysis of exponential and logarithmic real functions. These maps are used in many examples throughout the book. The chapter concludes with a brief discussion of the theory of concave functions on the real line.

1 Elements of Set Theory

1.1 Sets

Intuitively speaking, a set is a collection of objects.1 The distinguishing feature of a set is that whereas it may contain numerous objects, it is nevertheless conceived as a single entity. In the words of Georg Cantor, the great founder of abstract set theory, a set is a Many which allows itself to be thought of as a One. It is amazing how much follows from this simple idea.

The objects that a set S contains are called the elements (or members) of S. Clearly, to know S, it is necessary and sufcient to know all elements of S. The principal concept of set theory, then, is the relation of being an element/member of. The universally accepted symbol for this relation is ; that is, x S (or S x) means that x is an element of S (also read x is a member of S, or x is contained in S, or x belongs to S, or x is in S, or S includes x, etc.). We often write x, y S to denote that both x S and y S hold. For any natural number m, a statement like x1, . . . , xm S (or equivalently, xi S, i = 1, . . . , m) is understood analogously. If x S is a false statement, then we write x / S, and read x is not an element of S.

If the sets A and B have exactly the same elements, that is, x A iff x B, then we say that A and B are identical, and write A = B; otherwise we write A = B. 2 (So, for instance, {x, y} = {y, x}, {x, x} = {x}, and {{x}} = {x}.) If every member of A is also a member of B, then we say that A is a subset of B (also read A is a set in B or A is contained in B) and write A B (or B A). Clearly, A = B holds iff both A B and B A hold. If A B

1 The notion of object is left undened, that is, it can be given any meaning. All I demand

of our objects is that they be logically distinguishable. That is, if x and y are two objects,

x = y and x = y cannot hold simultaneously, and the statement either x = y or x = y is

a tautology.

2 Reminder. iff = if and only if.

1 Elements of Set Theory | 5

but A = B, then A is said to be a proper subset of B, and we denote this situation by writing A B (or B A).

For any set S that contains nitely many elements (in which case we say S is nite), we denote by |S| the total number of elements that S contains, and refer to this number as the cardinality of S. We say that S is a singleton if |S| = 1. If S contains innitely many elements (in which case we say S is innite), then we write |S| = . Obviously, we have |A| |B| whenever A B, and if A B and |A| < , then |A| < |B| .

We sometimes specify a set by enumerating its elements. For instance, {x, y, z} is the set that consists of the objects x, y, and z. The contents of the sets {x1, . . . , xm} and {x1, x2, . . .} are similarly described. For example, the set N of positive integers can be written as {1, 2, . . .}. Alternatively, one may describe a set S as a collection of all objects x that satisfy a given property P. If P(x) stands for the (logical) statement x satises the property P, then we can write S = {x : P(x) is a true statement}, or simply S = {x : P(x)}. If A is a set and B is the set that contains all elements x of A such that P(x) is true, we write B = {x A : P(x)}. For instance, where R is the set of all real numbers, the collection of all real numbers greater than or equal to 3 can be written as {x R : x 3}.

The symbol denotes the empty set, that is, the set that contains no elements (i.e., || = 0). Formally speaking, we can dene as the set {x : x = x}, for this description entails that x is a false statement for any object x. Consequently, we write

:= {x : x = x},

meaning that the symbol on the left-hand side is dened by that on the right-hand side.3 Clearly, we have S for any set S, which in particular implies that is unique. (Why?) If S = , we say that S is nonempty. For instance, {} is a nonempty set. Indeed, {} = the former, after all, is a set of sets that contains the empty set, while contains nothing. (An empty box is not the same thing as nothing!)

We dene the class of all subsets of a given set S as

2S := {T : T S},

3 Recall my notational convention: For any symbols and , either one of the expressions := and =: means that is dened by .


which is called the power set of S. (The choice of notation is motivated by the fact that the power set of a set that contains m elements has exactly 2m

elements.) For instance, 2 = {}, 22 = {, {}}, and 222 = {, {}, {{}}, {, {}}}, and so on.

Notation. Throughout this text, the class of all nonempty nite subsets of any given set S is denoted by P(S), that is,

P(S) := {T : T S and 0 < |T | < }. Of course, if S is nite, then P(S) = 2S\{}.

Given any two sets A and B, by A B we mean the set {x : x A or x B}, which is called the union of A and B. The intersection of A and B, denoted as A B, is dened as the set {x : x A and x B}. If A B = , we say that A and B are disjoint. Obviously, if A B, then A B = B and A B = A. In particular, S = S and S = for any set S.

Taking unions and intersections are commutative operations in the sense that

A B = B A and A B = B A for any sets A and B. They are also associative, that is,

A (B C) = (A B) C and A (B C) = (A B) C, and distributive, that is,

A (B C) = (A B) (A C) and A (B C) = (A B) (A C), for any sets A, B, and C.

Exercise 1 Prove the commutative, associative, and distributive laws of set theory stated above.

Exercise 2 Given any two sets A and B, by A\Bthe difference between A and Bwe mean the set {x : x A and x / B}. (a) Show that S\ = S, S\S = , and \S = for any set S. (b) Show that A\B = B\A iff A = B for any sets A and B. (c) (De Morgan Laws) Prove: For any sets A, B, and C,

A\(B C) = (A\B) (A\C) and A\(B C) = (A\B) (A\C).


Throughout this text we use the terms class or family only to refer to a nonempty collection of sets. So if A is a class, we understand that A = and that any member A A is a set (which may itself be a collection of sets). The union of all members of this class, denoted as A, or {A : A A}, or AAA, is dened as the set {x : x A for some A A}. Similarly, the intersection of all sets in A, denoted as A, or {A : A A}, or AAA, is dened as the set {x : x A for each A A}.

A common way of specifying a class A of sets is by designating a set I as a set of indices and by dening A := {Ai : i I}. In this case, A may be denoted as iIAi. If I = {k, k + 1, . . . , K} for some integers k and K with k < K , then we often write iK =kAi (or Ak AK ) for iIAi. Similarly, if I = {k, k + 1, . . .} for some integer k, then we may write Aii=k(or Ak Ak+1 ) for iIAi. Furthermore, for brevity, we frequently denote iK =1Ai as KAi, and i=1Ai as Ai, throughout the text. Similar notational conventions apply to intersections of sets as well.

Warning. The symbols and are left undened (in much the same way that the symbol 00 is undened in number theory).

Exercise 3 Let A be a set and B a class of sets. Prove that A B = {A B : B B} and A B = {A B : B B},

while

A\ B = {A\B : B B} and A\ B = {A\B : B B}.

A word of caution may be in order before we proceed further. While duly intuitive, the set theory we have outlined so far provides us with no demarcation criterion for identifyingwhat exactly constitutes a set. Thismay suggest that one is completely free in deeming any given collection of objects a set. But in fact, this would be a pretty bad idea that would entail serious foundational difculties. The best known example of such difculties was given by Bertrand Russell in 1902 when he asked if the set of all objects that are not members of themselves is a set: Is S := {x : x / x} a set?4 There is

4 While a bit unorthodox, x x may well be a statement that is true for some objects. For instance, the collection of all sets that I have mentioned in my life, say x, is a set that I have just mentioned, so x x. But the collection of all cheesecakes I have eaten in my life, say y, is not a cheesecake, so y / y.


nothing in our intuitive discussion above that forces us to conclude that S is not a set; it is a collection of objects (sets in this case) that is considered as a single entity. But we cannot accept S as a set, for if we do, we have to be able to answer the question, Is S S? If the answer is yes, then S S, but this implies S / S by denition of S. If the answer is no, then S / S, but this implies S S by denition of S. That is, we have a contradictory state of affairs no matter what! This is the so-called Russells paradox, which started a severe foundational crisis for mathematics that eventually led to a complete axiomatization of set theory in the early twentieth century.5

Roughly speaking, this paradox would arise only if we allowed unduly large collections to be qualied as sets. In particular, it will not cause any harm for the mathematical analysis that will concern us here, precisely because in all of our discussions, we will x a universal set of objects, say X , and consider sets like {x X : P(x)}, where P(x) is an unambiguous logical statement in terms of x. (We will also have occasion to work with sets of such sets, and sets of sets of such sets, and so on.) Once such a domain X is xed, Russells paradox cannot arise. Why, you may ask, cant we have the same problem with the set S := {x X : x / x}? No, because now we can answer the question: Is S S? The answer is no! The statement S S is false, simply because S / X . (For, if S X was the case, then we would end up with the contradiction S S iff S / S.)

So when the context is clear (that is, when a universe of objects is xed), and when we dene our sets as just explained, Russells paradox will not be a threat against the resulting set theory. But can there be any other paradoxes? Well, there is really not an easy answer to this. To even discuss the matter unambiguously, we must leave our intuitive understanding of the notion of set and address the problem through a completely axiomatic approach (in which we would leave the expression x S undened and give meaning to it only through axioms). This is, of course, not at all the place to do this. Moreover, the intuitive set theory that we covered here is more than enough for the mathematical analysis to come. We thus leave this topic by

5 Russells paradox is a classic example of the dangers of using self-referential statements carelessly. Another example of this form is the ancient paradox of the liar : Everything I say is false. This statement can be declared neither true nor false! To get a sense of some other kinds of paradoxes and the way axiomatic set theory avoids them, you might want to read the popular account of Rucker (1995).


referring the reader who wishes to get a broader introduction to abstract set theory to Chapter 1 of Schechter (1997) or Marek and Mycielski (2001); both of these expositions provide nice introductory overviews of axiomatic set theory. If you want to dig deeper, then try the rst three chapters of Enderton (1977).

1.2 Relations

An ordered pair is an ordered list (a, b) consisting of two objects a and b. This list is ordered in the sense that, as a dening feature of the notion of ordered pair, we assume the following: For any two ordered pairs (a, b) and (a , b), we have (a, b) = (a , b) iff a = a and b = b . 6

The (Cartesian) product of two nonempty sets A and B, denoted as A B, is dened as the set of all ordered pairs (a, b) where a comes from A and b comes from B. That is,

A B := {(a, b) : a A and b B}.

As a notational convention, we often write A2 for A A. It is easily seen that taking the Cartesian product of two sets is not a commutative operation. Indeed, for any two distinct objects a and b, we have {a} {b} ={(a, b)} = {(b, a)} = {b} {a}. Formally speaking, it is not associative either, for (a, (b, c)) is not the same thing as ((a, b), c). Yet there is a natural correspondence between the elements of A (B C) and (A B) C, so one can really think of these two sets as the same, thereby rendering the status of the set A B C unambiguous.7 This prompts us to dene an n-vector

6 This denes the notion of ordered pair as a new primitive for our set theory, but in fact, this is not really necessary. One can dene an ordered pair by using only the concept of set as (a, b) := {{a}, {a, b}}. With this denition, which is due to Kazimierz Kuratowski, one can prove that, for any two ordered pairs (a, b) and (a , b), we have (a, b) = (a , b) iff a = a and b = b . The if part of the claim is trivial. To prove the only if part, observe that (a, b) = (a , b) entails that either {a} = {a} or {a} = {a , b}. But the latter equality may hold only if a = a = b, so we have a = a in all contingencies. Therefore, (a, b) = (a , b) entails that either {a, b} = {a} or {a, b} = {a, b}. The latter case is possible only if b = b, while the former possibility arises only if a = b. But if a = b, then we have {{a}} = (a, b) = (a, b) = {{a}, {a, b}}, which holds only if {a} = {a, b}, that is, b = a = b .

Quiz. (Wiener) Show that we would also have (a, b) = (a , b) iff a = a and b = b, if we instead dened (a, b) as {{, {a}}, {{b}}}. 7 What is this natural correspondence?


(for any natural number n) as a list (a1, . . . , an), with the understanding that (a1, . . . , an) = (a1 , . . . , a n) iff ai = ai for each i = 1, . . . , n. The (Cartesian) product of n sets A1, . . . , An, is then dened as

A1 . . . An := {(a1, . . . , an) : ai Ai, i = 1, . . . , n}.

We often write XnAi to denote A1 An, and refer to XnAi as the n-fold product of A1, . . . , An. If Ai = S for each i, we thenwrite Sn for A1 . . .An, that is, Sn := XnS.

Exercise 4 For any sets A, B, and C, prove that

A(B C) = (AB) (AC) and A(B C) = (AB) (AC).

Let X and Y be two nonempty sets. A subset R of X Y is called a (binary) relation from X to Y . If X = Y , that is, if R is a relation from X to X , we simply say that it is a relation on X . Put differently, R is a relation on X iff R X 2. If (x, y) R, then we think of R as associating the object x with y, and if {(x, y), (y, x)} R = , we understand that there is no connection between x and y as envisaged by R. In concert with this interpretation, we adopt the convention of writing xRy instead of (x, y) R throughout this text.

Denition A relation R on a nonempty set X is said to be reexive if xRx for each x X , and complete if either xRy or yRx holds for each x, y X . It is said to be symmetric if, for any x, y X , xRy implies yRx, and antisymmetric if, for any x, y X , xRy and yRx imply x = y. Finally, we say that R is transitive if xRy and yRz imply xRz for any x, y, z X .

The interpretations of these properties are straightforward, so we do not elaborate on them here. But note: While every complete relation is reexive, there are no other logical implications between these properties.

Exercise 5 Let X be a nonempty set, and R a relation on X . The inverse of R is dened as the relation R1 := {(x, y) X 2 : yRx}.


(a) If R is symmetric, does R1 have to be also symmetric?

Antisymmetric? Transitive?

(b) Show that R is symmetric iff R = R1 . (c) If R1 and R2 are two relations on X , the composition of R1 and R2 is

the relation R2 R1 := {(x, y) X 2 : xR1z and zR2y for some z X }. Show that R is transitive iff R R R.

Exercise 6 A relation R on a nonempty set X is called circular if xRz and zRy imply yRx for every x, y, z X . Prove that R is reexive and circular iff it is reexive, symmetric, and transitive.

Exercise 7 H Let R be a reexive relation on a nonempty set X . The asymmetric part of R is dened as the relation PR on X as xPRy iff xRy but not yRx. The relation IR := R\PR on X is then called the symmetric part of R. (a) Show that IR is reexive and symmetric. (b) Show that PR is neither reexive nor symmetric. (c) Show that if R is transitive, so are PR and IR.

Exercise 8 Let R be a relation on a nonempty set X . Let R0 = R, and for each positive integer m, dene the relation Rm on X by xRmy iff there exist z1, . . . , zm X such that xRz1, z1Rz2, . . . , and zmRy. The relation tr(R) := R0 R1 is called the transitive closure of R. Show that tr(R) is transitive, and if R is a transitive relation with R R , then tr(R) R .

1.3 Equivalence Relations

In mathematical analysis, one often needs to identify two distinct objects when they possess a particular property of interest. Naturally, such an identication scheme should satisfy certain consistency conditions. For instance, if x is identied with y, then y must be identied with x. Similarly, if x and y are deemed identical, and so are y and z, then x and z should be identied. Such considerations lead us to the notion of equivalence relation.


Denition A relation on a nonempty set X is called an equivalence relation if it is reexive, symmetric, and transitive. For any x X , the equivalence class of x relative to is dened as the set

[x] := {y X : y x}. The class of all equivalence classes relative to , denoted as X /, is called the quotient set of X relative to , that is,

X / := {[x] : x X }.

Let X denote the set of all people in the world. Being a sibling of is an equivalence relation on X (provided that we adopt the convention of saying that any person is a sibling of himself). The equivalence class of a person relative to this relation is the set of all of his or her siblings. On the other hand, you would probably agree that being in love with is not an equivalence relation on X . Here are some more examples (that t better with the serious tone of this course).

Example 1 [1] For any nonempty set X , the diagonal relation DX := {(x, x) : x

X } is the smallest equivalence relation that can be dened on X (in the sense that if R is any other equivalence relation on X , we have DX R). Clearly, [x] = {x} for each x X . 8 At the other extreme is X 2 whichDX is the largest equivalence relation that can be dened on X . We have [x] = X for each x X .X 2

[2] By Exercise 7, the symmetric part of any reexive and transitive relation on a nonempty set is an equivalence relation.

[3] Let X := {(a, b) : a, b {1, 2, . . .}}, and dene the relation on X by (a, b) (c, d) iff ad = bc. It is readily veried that is an equivalence relation on X , and that [(a, b)] = (c, d) X : c = a ford b each (a, b) X .

[4] Let X := {. . . , 1, 0, 1, . . .}, and dene the relation on X by x y iff 12 (x y) X . It is easily checked that is an equivalence relation

8 I say an equally suiting name for DX is the equality relation. What do you think?


on X . Moreover, for any integer x, we have x y iff y = x 2m for some m X , and hence [x] equals the set of all even integers if x is even, and that of all odd integers if x is odd.

One typically uses an equivalence relation to simplify a situation in a way that all things that are indistinguishable from a particular perspective are put together in a set and treated as if they were a single entity. For instance, suppose that for some reason we are interested in the signs of people. Then, any two individuals who are of the same sign can be thought of as identical, so instead of the set of all people in the world, we would rather work with the set of all Capricorns, all Virgos, and so on. But the set of all Capricorns is of course none other than the equivalence class of any given Capricorn person relative to the equivalence relation of being of the same sign. So when someone says a Capricorn is. . . , then one is really referring to a whole class of people. The equivalence relation of being of the same sign divides the world into twelve equivalence classes, and we can then talk as if there were only twelve individuals in our context of reference.

To take another example, ask yourself how you would dene the set of positive rational numbers, given the set of natural numbers N := {1, 2, . . .}and the operation of multiplication. Well, you may say, a positive rational number is the ratio of two natural numbers. But wait, what is a ratio? Let us be a bit more careful about this. A better way of looking at things is to say that a positive rational number is an ordered pair (a, b) N2 , although in daily practice, we write ab instead of (a, b). Yet we dont want to say that each ordered pair in N2 is a distinct rational number. (We would like to think of 12 and

24 as the same number, for instance.) So we iden

tify all those ordered pairs that we wish to associate with a single rational number by using the equivalence relation introduced in Example 1.[3], and then dene a rational number simply as an equivalence class [(a, b)]. Of course, when we talk about rational numbers in daily practice, we simply talk of a fraction like 12 , not [(1, 2)], even though, formally speaking, what we really mean is [(1, 2)]. The equality 1 = 42 is obvious, pre2 cisely because the rational numbers are constructed as equivalence classes such that (2, 4) [(1, 2)].

This discussion suggests that an equivalence relation can be used to decompose a grand set of interest into subsets such that the members of


the same subset are thought of as identical, while the members of distinct subsets are viewed as distinct. Let us now formalize this intuition. By a partition of a nonempty set X , we mean a class of pairwise disjoint, nonempty subsets of X whose union is X . That is, A is a partition of X iff A 2X \{}, A = X and A B = for every distinct A and B in A. The next result says that the set of equivalence classes induced by any equivalence relation on a set is a partition of that set.

Proposition 1 For any equivalence relation on a nonempty set X , the quotient set X / is a partition of X .

Proof Take any nonempty set X and an equivalence relation on X . Since is reexive, we have x [x] for each x X . Thus any member of X / is nonempty, and {[x] : x X } = X . Now suppose that [x] [y] = for some x, y X . We wish to show that [x] = [y]. Observe rst that [x] [y] = implies x y. (Indeed, if z [x] [y], then x z and z y by symmetry of , so we get x y by transitivity of .) This implies that [x] [y], because if w [x], then w x (by symmetry of ), and hence w y by transitivity of . The converse containment is proved analogously.

The following exercise shows that the converse of Proposition 1 also holds. Thus the notions of equivalence relation and partition are really two different ways of looking at the same thing.

Exercise 9 Let A be a partition of a nonempty set X , and consider the relation on X dened by x y iff {x, y} A for some A A. Prove that is an equivalence relation on X .

1.4 Order Relations

Transitivity property is the dening feature of any order relation. Such relations are given various names depending on the properties they possess in addition to transitivity.


Denition A relation on anonempty set X is called a preorder on X if it is transitive and reexive. It is said to be a partial order on X if it is an antisymmetric preorder on X . Finally, is called a linear order on X if it is a partial order on X that is complete.

By a preordered set we mean a list (X , ) where X is a nonempty set and is a preorder on X . If is a partial order on X , then (X , ) is called a poset (short for partially ordered set), and if is a linear order on X , then (X , ) is called either a chain or a loset (short for linearly ordered set).

It is convenient to talk as if a preordered set (X , ) were indeed a set when referring to properties that apply only to X . For instance, by a nite preordered set, we understand a preordered set (X , ) with |X | < . Or, when we say that Y is a subset of the preordered set (X , ), we mean simply that Y X . A similar convention applies to posets and losets as well. Notation. Let (X , ) be a preordered set. Unless otherwise is stated explicitly, we denote by the asymmetric part of and by the symmetric part of (Exercise 7).

The main distinction between a preorder and a partial order is that the former may have a large symmetric part, while the symmetric part of the latter must equal the diagonal relation. As we shall see, however, in most applications this distinction is immaterial.

Example 2 [1] For any nonempty set X , the diagonal relation DX := {(x, x) : x

X } is a partial order on X . In fact, this relation is the only partial order on X that is also an equivalence relation. (Why?) The relation X 2, on the other hand, is a complete preorder, which is not antisymmetric unless X is a singleton.

[2] For any nonempty set X , the equality relation = and the subsethood relation are partial orders on 2X . The equality relation is not linear, and is not linear unless X is a singleton.

[3] (Rn , ) is a poset for any positive integer n, where is dened coordinatewise, that is, (x1, . . . , xn) (y1, . . . , yn) iff xi yi for each


i = 1, . . . , n. When we talk of Rn without specifying explicitly an alternative order, we always have in mind this partial order (which is sometimes called the natural (or canonical) order of Rn). Of course, (R, ) is a loset.

[4] Take any positive integer n and preordered sets (Xi, i), i = 1, . . . , n. The product of the preordered sets (Xi, i), denoted as n(Xi, i), is the preordered set (X , ) with X := XnXi and

(x1, . . . , xn) (y1, . . . , yn) iff xi i yi for all i = 1, . . . , n. In particular, (Rn , ) = n(R, ).

Example 3 In individual choice theory, a preference relation on a nonempty alternative set X is dened as a preorder on X . Here the reexivity is a trivial condition to require, and transitivity is viewed as a fundamental rationality postulate. (We will talk more about this in Section B.4.) The strict preference relation is dened as the asymmetric part of (Exercise 7). This relation is transitive but not reexive. The indifference relation is then dened as the symmetric part of , and is easily checked to be an equivalence relation on X . For any x X , the equivalence class [x] is called in this context the indifference class of x, and is simply a generalization of the familiar concept of the indifference curve that passes through x. In particular, Proposition 1 says that no two distinct indifference sets can have a point in common. (This is the gist of the fact that distinct indifference curves cannot cross!)

In social choice theory, one often works with multiple (complete) preference relations on a given alternative set X . For instance, suppose that there are n individuals in the population, and i stands for the preference relation of the ith individual. The Pareto dominance relation on X is dened as x y iff x i y for each i = 1, . . . , n. This relation is a preorder on X in general, and a partial order on X if each i is antisymmetric.

Let (X , ) be a preordered set. By an extension of we understand a preorder on X such that and , where is the asymmetric part of . Intuitively speaking, an extension of a preorder is more complete than the original relation in the sense that it allows one to compare more elements, but it certainly agrees exactly with the original relation when


the latter applies. If is a partial order, then it is an extension of iff . (Why?)

A fundamental result of order theory says that every partial order can be extended to a linear order, that is, for every poset (X , ) there is a loset (X , ) with . Although it is possible to prove this by mathematical induction when X is nite, the proof in the general case is built on a relatively advanced method that we will cover later in the course. Relegating its proof to Section 1.7, we only state here the result for future reference.9

Szpilrajns Theorem Every partial order on a nonempty set X can be extended to a linear order

on X .

A natural question is whether the same result holds for preorders as well. The answer is yes, and the proof follows easily from Szpilrajns Theorem by means of a standard method.

Corollary 1 Let (X , ) be a preordered set. There exists a complete preorder on X that extends .

Proof Let denote the symmetric part of , which is an equivalence relation. Then (X /, ) is a poset where is dened on X / by

[x] [y] if and only if x y. By Szpilrajns Theorem, there exists a linear order on X / such that . We dene on X by

x y if and only if [x] [y]. It is easily checked that is a complete preorder on X with and , where and are the asymmetric parts of and , respectively.

9 For an extensive introduction to the theory of linear extensions of posets, see Bonnet and Pouzet (1982).


Exercise 10 Let (X , ) be a preordered set, and dene L() as the set of all complete preorders that extend . Prove that = L(). (Where do you use Szpilrajns Theorem in the argument?)

Exercise 11 Let (X , ) be a nite preordered set. Taking L() as in the previous exercise, we dene dim(X , ) as the smallest positive integer k such that = R1 Rk for some Ri L(), i = 1, . . . , k. (a) Show that dim(X , ) X 2 . (b) What is dim(X , DX )? What is dim(X , X 2)? (c) For any positive integer n, show that dim(n(Xi, i)) = n, where

(Xi, i) is a loset with |Xi| 2 for each i = 1, . . . , n. (d) Prove or disprove: dim(2X , ) = |X | .

Denition Let (X , ) be a preordered set, and = Y X . An element x of Y is said to be -maximal in Y if there is no y Y with y x, and -minimal in Y if there is no y Y with x y. If x y for all y Y , then x is called the -maximum of Y , and if y x for all y Y , then x is called the -minimum of Y .

Obviously, for any preordered set (X , ), every -maximum of a nonempty subset of X is -maximal in that set. Also note that if (X , ) is a poset, then there can be at most one -maximum of any Y 2X \{}.

Example 4 [1] Let X be any nonempty set, and = Y X . Every element of Y

is both DX -maximal and DX -minimal in Y . Unless it is a singleton, Y has neither a DX -maximum nor a DX -minimum element. On the other hand, every element of Y is both X 2-maximum and X 2-minimum of Y .

[2] Given any nonempty set X , consider the poset (2X , ), and take any nonempty A 2X . The class A has a -maximum iff A A, and it has a -minimum iff A A. In particular, the -maximum of 2X is X and the -minimum of 2X is .

[3] (Choice Correspondences) Given a preference relation on an alternative set X (Example 3) and a nonempty subset S of X , we dene the set of choices from S for an individual whose preference relation is


as the set of all -maximal elements in S. That is, denoting this set as C(S), we have

C(S) := {x S : y x for no y S}. Evidently, if S is a nite set, then C(S) is nonempty. (Proof?) Moreover, if S is nite and is complete, then there exists at least one -maximum element in S. The niteness requirement cannot be omitted in this statement, but as we shall see throughout this book, there are various ways in which it can be substantially weakened.

Exercise 12 (a) Which subsets of the set of positive integers have a -minimum?

Which ones have a -maximum? (b) If a set in a poset (X , ) has a unique -maximal element, does

that element have to be a -maximum of the set? (c) Which subsets of a poset (X , ) possess an element that is both

-maximum and -minimum? (d) Give an example of an innite set in R2 that contains a unique

-maximal element that is also the unique -minimal element of the set.

Exercise 13H Let be a complete relation on a nonempty set X , and S a nonempty nite subset of X . Dene

c(S) := {x S : x y for all y S}. (a) Show that c(S) = if is transitive. (b) We say that is acyclic if there does not exist a positive integer k

such that x1, . . . , xk X and x1 xk x1. Show that every transitive relation is acyclic, but not conversely.

(c) Show that c(S) = if is acyclic. (d) Show that if c(T ) = for every nite T 2X \{}, then must be

acyclic.

Exercise 14H Let (X , ) be a poset, and take any Y 2X \{} that has a -maximal element, say x . Prove that can be extended to a linear order on X such that x is -maximal in Y .

Exercise 15 Let (X , ) be a poset. For any Y X , an element x in X is said to be an -upper bound for Y if x y for all y Y ; a -lower bound


for Y is dened similarly. The -supremum of Y , denoted sup Y , is dened as the -minimum of the set of all -upper bounds for Y , that is, sup Y is an -upper bound for Y and has the property that z sup Y for any -upper bound z for Y . The -inmum of Y , denoted as inf Y , is dened analogously. (a) Prove that there can be only one -supremum and only one

-inmum of any subset of X . (b) Show that x y iff sup{x, y} = x and inf {x, y} = y, for any

x, y X . (c) Show that if sup X X (that is, if sup X exists), then

inf = sup X .

(d) If is the diagonal relation on X , and x and y are any two distinct

members of X , does sup{x, y} exist? (e) If X := {x, y, z, w} and := {(z, x), (z, y), (w, x), (w, y)}, does

sup{x, y} exist?

Exercise 16H Let (X , ) be a poset. If sup{x, y} and inf {x, y} exist for all x, y X , then we say that (X , ) is a lattice. If sup Y and inf Y exist for all Y 2X , then (X , ) is called a complete lattice. (a) Show that every complete lattice has an upper and a lower bound. (b) Show that if X is nite and (X , ) is a lattice, then (X , ) is a

complete lattice.

(c) Give an example of a lattice which is not complete. (d) Prove that (2X , ) is a complete lattice. (e) Let X be a nonempty subset of 2X such that X X and A X for

any (nonempty) class A X . Prove that (X , ) is a complete lattice.

1.5 Functions

Intuitively, we think of a function as a rule that transforms the objects in a given set to those of another. Although this is not a formal denitionwhat is a rule?we may now use the notion of binary relation to formalize the idea. Let X and Y be any two nonempty sets. By a function f that maps X into Y , denoted as f : X Y , we mean a relation f X Y such that

(i) for every x X , there exists a y Y such that x f y; (ii) for every y, z Y with x f y and x f z, we have y = z.


Here X is called the domain of f and Y the codomain of f . The range of f is, on the other hand, dened as

f (X ) := {y Y : x f y for some x X }. The set of all functions that map X into Y is denoted by YX . For instance, {0, 1}X is the set of all functions on X whose values are either 0 or 1, and R[0,1] is the set of all real-valued functions on [0, 1]. The notation f YX will be used interchangeably with the expression f : X Y throughout this course. Similarly, the term map is used interchangeably with the term function.

Although our denition of a function may look a bit strange at rst, it is hardly anything other than a set-theoretic formulation of the concept we use in daily discourse. After all, we want a function f thatmaps X into Y to assign each member of X to a member of Y , right? Our denition says simply that one can think of f as a set of ordered pairs, so (x, y) f means x is mapped to y by f . Put differently, all that f does is completely identied by the set {(x, f (x)) X Y : x X }, which is what f is. The familiar notation f (x) = y (which we shall also adopt in the rest of the exposition) is then nothing but an alternative way of expressing x f y. When f (x) = y, we refer to y as the image (or value) of x under f . Condition (i) says that every element in the domain X of f has an image under f in the codomain Y . In turn, condition (ii) states that no element in the domain of f can have more than one image under f .

Some authors adhere to the intuitive denition of a function as a rule that transforms one set into another and refer to the set of all ordered pairs (x, f (x)) as the graph of the function. Denoting this set by Gr( f ), then, we can write

Gr( f ) := {(x, f (x)) X Y : x X }. According to the formal denition of a function, f and Gr( f ) are the same thing. So long as we keep this connection in mind, there is no danger in thinking of a function as a rule in the intuitive way. In particular, we say that two functions f and g are equal if they have the same graph, or equivalently, if they have the same domain and codomain, and f (x) = g(x) for all x X . In this case, we simply write f = g .

If its range equals its codomain, that is, if f (X ) = Y , then one says that f maps X onto Y , and refers to it as a surjection (or as a surjective


function/map). If f maps distinct points in its domain to distinct points in its codomain, that is, if x = y implies f (x) = f (y) for all x, y X , then we say that f is an injection (or a one-to-one or injective function/map). Finally, if f is both injective and surjective, then it is called a bijection (or a bijective function/map). For instance, if X := {1, . . . , 10}, then f := {(1, 2), (2, 3), . . . , (10, 1)} is a bijection in XX , while g XX , dened as g(x) := 3 for all x X , is neither an injection nor a surjection. When considered as a map in ({0} X )X , f is an injection but not a surjection. Warning. Every injective function can be viewed as a bijection, provided that one views the codomain of the function as its range. Indeed, if f : X Y is an injection, then the map f : X Z is a bijection, where Z := f (X ). This is usually expressed as saying that f : X f (X ) is a bijection.

Before we consider some examples, let us note that a common way of deningaparticular function in agiven context is todescribe thedomainand codomain of that function and the image of a generic point in the domain. So one would say something like, let f : X Y be dened by f (x) := . . . or consider the function f YX dened by f (x) := . . . . For example, by the function f : R R+ dened by f (t) := t2, we mean the surjection that transforms every real number t to the nonnegative real number t2. Since the domain of the function is understood from the expression f : X Y (or f YX ), it is redundant to add the phrase for all x X after the expression f (x) := . . . , although sometimes we may do so for clarity. Alternatively, when the codomain of the function is clear, a phrase like the map x f (x) on X is commonly used. For instance, one may refer to the quadratic function mentioned above unambiguously as the map t t2 on R.

Example 5 In the following examples, X and Y stand for arbitrary nonempty sets.

[1] A constant function is one that assigns the same value to every element of its domain, that is, f YX is constant iff there exists a y Y such that f (x)= y for all x X . (Formally speaking, this constant function is the set X {y}.) Obviously, f (X )={y} in this case, so a constant function is not surjective unless its codomain is a singleton, and it is not injective unless its domain is a singleton.


[2] A function whose domain and codomain are identical, that is, a function in XX , is called a self-map on X . An important example of a self-map is the identity function on X . This function is denoted as idX , and it is dened as idX (x) := x for all x X . Obviously, idX is a bijection, and formally speaking, it is none other than the diagonal relation DX .

[3] Let S X . The function that maps X into {0, 1} such that every member of S is assigned to 1 and all the other elements of X are assigned to zero is called the indicator function of S in X . This function is denoted as 1S (assuming that the domain X is understood from the context). By denition, we have

1, if x S1S(x) := .

0, if x X \S You can check that, for every A, B X , we have 1AB + 1AB = 1A + 1B and 1AB = 1A1B.

The following examples point to some commonly used methods of obtaining new functions from a given set of functions.

Example 6 In the following examples, X , Y , Z, and W stand for arbitrary nonempty sets.

[1] Let Z X W , and f YX . By the restriction of f to Z, denoted as f |Z , we mean the function f |Z YZ dened by f |Z (z) := f (z). By an extension of f to W , on the other hand, we mean a function f YW with f |X = f , that is, f (x) = f (x) for all x X . If f is injective, so must f |Z , but surjectivity of f does not entail that of f |Z . Of course, if f is not injective, f |Z may still turn out to be injective (e.g., t t2 is not injective on R, but it is so on R+).

[2] Sometimes it is possible to extend a given function by combining it with another function. For instance, we can combine any f YX and g WZ to obtain the function h : X Z Y W dened by

f (t), if t X h(t) := ,

g(t), if t Z


provided that X Z = , or X Z = and f |X Z = g |X Z . Note that this method of combining functions does not work if f (t) = g(t) for some t X Z. For, in that case h would not be well-dened as a function. (What would be the image of t under h?)

[3] A function f XX Y denedby f (x, y) := x is called the projection from X Y onto X . 10 (The projection from X Y onto Y is similarly dened.) Obviously, f (X Y ) = X , that is, f is necessarily surjective. It is not injective unless Y is a singleton.

[4] Given functions f : X Z and g : Z Y , we dene the composition of f and g as the function g f : X Y by g f (x) := g( f (x)). (For easier reading, we often write (g f )(x) instead of g f (x).) This denition accords with the way we dened the composition of two relations (Exercise 5). Indeed, we have (g f )(x) = {(x, y) : x f z and z g y for some z Z}.

Obviously, idZ f = f = f idX . Evenwhen X = Y = Z, the operation of taking compositions is not commutative. For instance, if the self-maps f and g on R are dened by f (t) := 2 and g(t) := t2, respectively, then ( g f )(t) = 4 and ( f g)(t) = 2 for any real number t. The composition operation is, however, associative. That is, h ( g f ) = (h g) f for all f YX , g ZY and h WZ .

Exercise 17 Let be an equivalence relation on a nonempty set X . Show that the map x [x] on X (called the quotient map) is a surjection on X which is injective iff = DX .

Exercise 18H (A Factorization Theorem) Let X and Y be two nonempty sets. Prove: For any function f : X Y , there exists a nonempty set Z, a surjection g : X Z, and an injection h : Z Y such that f = h g .

Exercise 19 Let X , Y , and Z be nonempty sets, and consider any f , g YX and u, v ZY . Prove: (a) If f is surjective and u f = v f , then u = v. (b) If u is injective and u f = u g , then f = g . (c) If f and u are injective (respectively, surjective), then so is u f .

10 Strictly speaking, I should write f ((x, y)) instead of f (x, y), but thats just splitting hairs.


Exercise 20H Show that there is no surjection of the form f : X 2X for any nonempty set X .

For any given nonempty sets X and Y , the (direct) image of a set A X under f YX , denoted f (A), is dened as the collection of all elements y in Y with y = f (x) for some x A. That is,

f (A) := { f (x) : x A}. The rangeof f is thus the imageof its entire domain: f (X ) = { f (x) : x X }. (Note. If f (A) = B, then one says that f maps A onto B.)

The inverse image of a set B in Y under f , denoted as f 1(B), is dened as the set of all x in X whose images under f belong to B, that is,

f 1(B) := {x X : f (x) B}. By convention, we write f 1(y) for f 1({y}), that is,

f 1(y) := {x X : f (x) = y} for any y Y . Obviously, f 1(y) is a singleton for each y Y iff f is an injection. For instance, if f stands for the map t t2 on R, then f 1(1) = {1, 1}, whereas f |1 (1) = {1}.

R+The issue of whether or not one can express the image (or the in

verse image) of a union/intersection of a collection of sets as the union/ intersection of the images (inverse images) of each set in the collection arises quite often in mathematical analysis. The following exercise summarizes the situation in this regard.

Exercise 21 Let X and Y be nonempty sets and f YX . Prove that, for any (nonempty) classes A 2X and B 2Y , we have

f (A) =

{ f (A) : A A} and f (A)

{ f (A) : A A}, whereas f 1 (B) = { f 1(B) : B B} f 1 (B) = { f 1(B) : B B}and .

A general rule that surfaces from this exercise is that inverse images are quite well-behaved with respect to the operations of taking unions and intersections, while the same cannot be said for direct images in the case of taking intersections. Indeed, for any f YX , we have f (A B) f (A) f (B) for


all A, B X if, and only if, f is injective.11 The if part of this assertion is trivial. The only if part follows from the observation that, if the claim was not true, then, for any distinct x, y X with f (x) = f (y), we would nd = f () = f ({x} {y}) = f ({x}) f ({y}) = { f (x)}, which is absurd.

Finally, we turn to the problem of inverting a function. For any function f YX , let us dene the set

f 1 := {(y, x) Y X : x f y}

which is none other than the inverse of f viewed as a relation (Exercise 5). This relation simply reverses the map f in the sense that if x is mapped to y by f , then f 1 maps y back to x. Now, f 1 may or may not be a function. If it is, we say that f is invertible and f 1 is the inverse of f . For instance, f : R R+ dened by f (t) := t2 is not invertible (since (1, 1) f 1 and (1, 1) f 1, that is, 1 does not have a unique image under f 1), whereas f | is invertible and f |1 (t) = t for all t R.R+ R+

The following result gives a simple characterization of invertible functions.

Proposition 2 Let X and Y be two nonempty sets. A function f YX is invertible if, and only if, it is a bijection.

Exercise 22 Prove Proposition 2.

By using the composition operation dened in Example 6.[4], we can give another useful characterization of invertible functions.

Proposition 3 Let X and Y be two nonempty sets. A function f YX is invertible if, and only if, there exists a function g X Y such that g f = idX and f g = idY .

11 Of course, this does not mean that f (A B) = f (A) f (B) can never hold for a function that is not one-to-one. It only means that, for any such function f , we can always nd nonempty sets A and B in the domain of f such that f (A B) f (A) f (B) is false.


Proof The only if part is readily obtained upon choosing g := f 1. To prove the if part, suppose there exists a g XY with g f = idX and f g = idY , and note that, by Proposition 2, it is enough to show that f is a bijection. To verify the injectivity of f , pick any x, y X with f (x) = f (y), and observe that

x = idX (x) = ( g f )(x) = g( f (x)) = g( f (y)) = ( g f )(y) = idX (y) = y. To see the surjectivity of f , take any y Y and dene x := g(y). Then we have

f (x) = f ( g(y)) = ( f g)(y) = idY (y) = y, which proves Y f (X ). Since the converse containment is trivial, we are done.

1.6 Sequences, Vectors, and Matrices

Bya sequence in a givennonempty set X , we intuitively mean an ordered array of the form (x1, x2, . . .)where each term xi of the sequence is amember of X . (Throughout this text we denote such a sequence by (xm), but note that some books prefer instead the notation (xm) 1.) As in the case of ordered pairs, m=one could introduce thenotionof a sequence as anewobject to our set theory, but again there is really no need to do so. Intuitively, we understand from the notation (x1, x2, . . .) that the ith term in the array is xi. But then we can think of this array as a function that maps the set N of positive integers into X in the sense that it tells us that the ith term in the array is xi by mapping i to xi. With this denition, our intuitive understanding of the ordered array (x1, x2, . . .) is formally captured by the function {(i, xi) : i = 1, 2, . . .} = f . Thus, we dene a sequence in a nonempty set X as any function f : N X , and represent this function as (x1, x2, . . .) where xi := f (i) for each i N. Consequently, the set of all sequences in X is equal to X N. As is common, however, we denote this set as X throughout the text.

By a subsequence of a sequence (xm) X , we mean a sequence that is made up of the terms of (xm) that appear in the subsequence in the same order they appear in (xm). That is, a subsequence of (xm) is of the form (xm1 , xm2 , . . .), where (mk) is a sequence in N such that m1 < m2 < . (We denote this subsequence as (xmk ).) Once again, we use the notion of


function to formalize this denition. Strictly speaking, a subsequence of a sequence f X N is a function of the form f , where : N N is strictly increasing (that is, (k) < (l) for any k, l N with k < l). We represent this function as the array (xm1 , xm2 , . . .) with the understanding that mk = (k) and xmk = f (mk) for each k = 1, 2, . . . . For instance, (xmk ) := (1, 31 , 51 , . . .) is a subsequence of (xm) := (m 1 ) R. Here (xm) is a representation for the function f RN, which is dened by f (i) := 1 i , and (xmk ) is a representation of the map f , where (k) := 2k 1 for each k N.

By a double sequence in X , we mean an innite matrix each term of which is a member of X . Formally, a double sequence is a function f X NN . As in the case of sequences, we represent this function as (xkl), with the understanding that xkl := f (k, l). The set of all double sequences in X equals X NN, but it is customary to denote this set as X . We note that one can always view (in more than one way) a double sequence in X as a sequence of sequences in X , that is, as a sequence in X . For instance, we can think of (xkl) as ((x1l), (x2l), . . .) or as ((xk1), (xk2), . . .).

The basic idea of viewing a string of objects as a particular function also applies to nite strings, of course. For instance, how about X {1,...,n}, where X is a nonempty set and n is some positive integer? The preceding discussion shows that this function space is none other than the set {(x1, . . . , xn) : xi X , i = 1, . . . , n}. Thus we may dene an n-vector in X as a function f : {1, . . . , n} X , and represent this function as (x1, . . . , xn) where xi := f (i) for each i = 1, . . . , n. (Check that (x1, . . . , xn) = (x1 , . . . , xn ) iff xi = xi for each i = 1, . . . , n, so everything is in concert with the way we dened n-vectors in Section 1.2.) The n-fold product of X is then dened as X {1,...,n}, but is denoted as Xn. (So Rn = R{1,...,n}. This makes sense, no?) The main lesson is that everything that is said about arbitrary functions also applies to sequences and vectors.

Finally, for any positive integers m and n, by an m n matrix (read m by n matrix) in a nonempty set X , we mean a function f : {1, . . . , m} {1, . . . , n} X . We represent this function as [aij]mn, with the understanding that aij := f (i, j) for each i = 1, . . . , m and j = 1, . . . , n. (As you know, one often views a matrix like [aij]mn as a rectangular array with m rows and n columns in which aij appears in the ith row and jth column.)

The set of all m n matrices in X is X {1,...,m}{1,...,n}, but it is much better to denote this set as Xmn. Needless to say, both X 1n and Xn1 can be identied with Xn. (Wait, what does this mean?)


1.7 A Glimpse of Advanced Set Theory: The Axiom of Choice

We now turn to a problem that we have so far conveniently avoided: How do we dene the Cartesian product of innitely many nonempty sets? Intuitively speaking, the Cartesian product of all members of a class A of sets is the set of all collections each of which contains one and only one element of each member of A. That is, a member of this product is really a function on A that selects a single element from each set in A. The question is simple to state: Does there exist such a function?

If |A| < , then the answer would obviously be yes, because we can construct such a function by choosing an element from each set in A one by one. But when A contains innitely many sets, then this method does not readily work, so we need to prove that such a function exists.

To get a sense of this, suppose A := {A1, A2, . . .}, where = Ai N for each i = 1, 2, . . . . Then were okay. We can dene f : A A by f (A) := the smallest element of A this well-denes f as a map that selects one element from each member of A simultaneously. Or, if each Ai is a bounded interval in R, then again were ne. This time we can dene f , say, as follows: f (A) := the midpoint of A. But what if all we knew was that each Ai consists of real numbers? Or worse, what if we were not told anything about the contents of A? You see, in general, we cant write down a formula, or an algorithm, the application of which yields such a function. Then how do you know that such a thing exists in the rst place?12

12 But, how about the following algorithm? Start with A1, and pick any a1 in A1. Now move to A2 and pick any a2 A2. Continue this way, and dene g : A A by g(Ai) = ai , i = 1, 2, . . .. Arent we done? No, we are not! The function at hand is not well-denedits denition does not tell me exactly which member of A27 is assigned to g(A27)this is very much unlike how I dened f above in the case where each Ai was contained in N (or was a bounded interval).

Perhaps you are still not quite comfortable about this. You might think that f is well-dened, its just that it is dened recursively. Let me try to illustrate the problem by means of a concrete example. Take any innite set S, and ask yourself if you can dene an injection f from N into S. Sure, you might say, recursion is again the name of the game. Let f (1) be any member a1 of S. Then let f (2) be any member of S\{a1}, f (3) any member S\{a1, a2}, and so on. Since S\T = for any nite T S, this well-denes f , recursively, as an injection from N into S. Wrong! If this were the case, on the basis of the knowledge of f (1), . . . , f (26), I would know the value of f at 27. The denition of f doesnt do thatit just points to some arbitrary member of A27so it is not a proper denition at all.

(Note. As obvious as it might seem, the proposition for any innite set S, there is an injection in SN, cannot be proved within the standard realm of set theory.)


In fact, it turns out that the problem of nding an f : A A for any given class A of sets cannot be settled in one way or another by means of the standard axioms of set theory.13 The status of our question is thus a bit odd, it is undecidable.

To make things a bit more precise, let us state formally the property that we are after.

The Axiom of Choice. For any (nonempty) class A of sets, there exists a function f : A A such that f (A) A for each A A.

One can reword this in a few other ways.

Exercise 23 Prove that the Axiom of Choice is equivalent to the following statements. (i) For any nonempty set S, there exists a function f : 2S\{} S

such that f (A) A for each = A S. (ii) (Zermelos Postulate) If A is a (nonempty) class of sets such that

A B = for each distinct A, B A, then there exists a set S such that |S A| = 1 for every A A.

(iii) For any nonempty sets X and Y , and any relation R from X into Y , there is a function f : Z Y with = Z X and f R. (That is: Every relation contains a function.)

The rst thing to note about the Axiom of Choice is that it cannot be disproved by using the standard axioms of set theory. Provided that these axioms are consistent (that is, no contradiction may be logically deduced from them), adjoining the Axiom of Choice to these axioms yields again a consistent set of axioms. This raises the possibility that perhaps the Axiom of Choice can be deduced as a theorem from the standard axioms. The second thing to know about the Axiom of Choice is that this is false, that is, theAxiomofChoice is not provable from the standard axiomsof set theory.14

13 For brevity, I am again being imprecise about this standard set of axioms (called the Zermelo-Fraenkel-Skolem axioms). For the present discussion, nothing will be lost if you just think of these as the formal properties needed to construct the set theory we outlined intuitively earlier. It is fair to say that these axioms have an unproblematic standing in mathematics. 14 These results are of extreme importance for the foundations of the entire eld of mathematics. The rst one was proved by Kurt Gdel in 1939 and the second one by Paul Cohen in 1963.


Weare thenat a crossroads. Wemust either reject the validity of theAxiom of Choice and conne ourselves to the conclusions that can be reached only on the basis of the standard axioms of set theory, or alternatively, adjoin the Axiom of Choice to the standard axioms to obtain a richer set theory that is able to yield certain results that could not have been proved within the connes of the standard axioms. Most analysts follow the second route. However, it is fair to say that the status of the Axiom of Choice is in general viewed as less appealing than the standard axioms, so one often makes it explicit if this axiom is a prerequisite for a particular theorem to be proved. Given our applied interests, we will be more relaxed about this matter and mention the (implicit) use of the Axiom of Choice in our arguments only rarely.

As an immediate application of the Axiom of Choice, we now dene the Cartesian product of an arbitrary (nonempty) class A of sets as the set of all f : A A with f (A) A for each A A. We denote this set by XA, and note that XA = because of the Axiom of Choice. If A = {Ai : i I}, where I is an index set, then we write XiIAi for XA. Clearly, XiIAi is the set of all maps f : I {Ai : i I} with f (i) Ai for each i I. It is easily checked that this denition is consistent with the denition of the Cartesian product of nitely many sets given earlier.

There are a few equivalent versions of the Axiom of Choice that are often more convenient to use in applications than the original statement of the axiom. To state the most widely used version, let us rst agree on some terminology. For any poset (X , ), by a poset in (X , ), we mean a poset like (Y , Y2) with Y X , but we denote this poset more succinctly as (Y , ). By an upper bound for such a poset, we mean an element x of X with x y for all y Y (Exercise 15).

Zorns Lemma If every loset in a given poset has an upper bound, then that poset must have

a maximal element.

Although this is a less intuitive statement than theAxiomofChoice (no?), it can in fact be shown to be equivalent to the Axiom of Choice.15 (That is, we can deduce Zorns Lemma from the standard axioms and the Axiom of Choice, and we can prove the Axiom of Choice by using the standard axioms

15 For a proof, see Enderton (1977, pp. 151153) or Kelley (1955, pp. 3235).


and Zorns Lemma.) Since we take the Axiom of Choice as true in this text, therefore, we must also accept the validity of Zorns Lemma.

We conclude this discussion by means of two quick applications that illustrate how Zorns Lemma is used in practice. We will see some other applications in later chapters.

Let us rst prove the following fact:

The Hausdorff Maximal Principle There exists a -maximal loset in every poset.

Proof Let (X , ) be a poset, and

L(X , ) := Z X : (Z, ) is a loset . (Observe that L(X , ) = by reexivity of .) We wish to show that there is a -maximal element of L(X , ). This will follow from Zorns Lemma, if we can show that every loset in the poset (L(X , ), ) has an upper bound, that is, for any A L(X , ) such that (A, ) is a loset, there is a member of L(X , ) that contains A. To establish that this is indeed the case, take any such A, and let Y := A. Then is a complete relation on Y , because, since linearly orders A, for any x, y Y we must have x, y A for some A A (why?), and hence, given that (A, ) is a loset, we have either x y or y x. Therefore, (Y , ) is a loset, that is, Y L(X , ). But it is obvious that Y A for any A A.

In fact, the Hausdorff Maximal Principle is equivalent to the Axiom of Choice.

Exercise 24 Prove Zorns Lemma assuming the validity of the Hausdorff Maximal Principle.

As another application of Zorns Lemma, we prove Szpilrajns Theorem.16 Our proof uses the Hausdorff Maximal Principle, but you now know that this is equivalent to invoking Zorns Lemma or the Axiom of Choice.

16 In case you are wondering, Szpilrajns Theorem is not equivalent to the Axiom of Choice.

2 Real Numbers | 33

Proof of Szpilrajns Theorem Let be a partial order on a nonempty set X . Let TX be the set of all partial orders on X that extend . Clearly, (TX , ) is a poset, so by the Hausdorff Maximal Principle, it has a maximal loset, say, (A, ). Dene := A. Since (A, ) is a loset, is a partial order on X that extends . (Why?) is in fact complete. To see this, suppose we can nd some x, y X with neither x y nor y x. Then the transitive closure of {(x, y)} is a member of TX that contains as a proper subset (Exercise 8). (Why exactly?) This contradicts the fact that (A, ) is a maximal loset within (TX , ). (Why?) Thus is a linear order, and we are done.

2 Real Numbers

This course assumes that the reader has a basic understanding of the real numbers, so our discussion here will be brief and duly heuristic. In particular, we will not even attempt to give a construction of the set R of real numbers. Instead we will mention some axioms that R satises, and focus on certain properties that R possesses. Some books on real analysis give a fuller view of the construction of R, some talk about it even less than we do. If you are really curious about this, its best if you consult a book that specializes in this sort of a thing. (Try, for instance, Chapters 4 and 5 of Enderton (1977).)

2.1 Ordered Fields

In this subsection we talk briey about a few topics in abstract algebra that will facilitate our discussion of real numbers.

Denition Let X be any nonempty set. We refer to a function of the form : X X X as a binary operation on X , and write x y instead of (x, y) for any x, y X .

For instance, the usual addition and multiplication operations + and are binary operations on the set N of natural numbers. The subtraction operation is, on the other hand, not a binary operation on N (e.g., 1+(2) / N), but it is a binary operation on the set of all integers.


Denition Let X be any nonempty set, let + and be two binary operations on X , and let us agree to write xy for x y for simplicity. The list (X , +, ) is called a eld if the following properties are satised:

(i) (Commutativity) x + y = y + x and xy = yx for all x, y X ; (ii) (Associativity) (x + y) + z = x + (y + z) and (xy)z = x(yz) for all

x, y, z X ;17

(iii) (Distributivity) x(y + z) = xy + xz for all x, y, z X ; (iv) (Existence of Identity Elements) There exist elements 0 and 1 in X

such that 0 + x = x = x + 0 and 1x = x = x1 for all x X ; (v) (Existence of Inverse Elements) For each x X there exists an element

x in X (the additive inverse of x) such that x +x = 0 = x + x, and for each x X \{0} there exists an element x1 in X (the multiplicative inverse of x) such that xx1 = 1 = x1x.

A eld (X , +, ) is an algebraic structure that envisions two binary operations, + and , on the set X in a way that makes a satisfactory arithmetic possible. In particular, given the + and operations, we can dene the two other (inverse) operations and / by x y := x +y and x/y := xy1, the latter provided that y = 0. (Strictly speaking, the division operation / is not a binary operation; for instance, 1/0 is not dened in X .)

Pretty much the entire arithmetic that we are familiar with in the context of R can be performed within an arbitrary eld. To illustrate this, let us establish a few arithmetic laws that you may recall from high school algebra. In particular, let us show that

x +y = x +z iff y = z, (x) = x and (x +y) = x +y (1) in any eld (X , +, ). The rst claim is a cancellation law, which is readily provedbyobserving that, for any w X , wehave w = 0+w = (x+x)+w = x + (x + w). Thus, x + y = x + z implies y = x + (x + y) = z, and were done. As an immediate corollary of this cancellation law, we nd that

17 Throughout this exposition, (w) is the same thing as w, for any w X . For instance, (x + y) corresponds to x + y, and (x) corresponds to x. The brackets are used at times only for clarity.

2 Real Numbers | 35

the additive inverse of each element in X is unique. (The same holds for the multiplicative inverses as well. Quiz. Prove!) On the other hand, the second claim in (1) is true because

x = x+0 = x+(x+(x)) = (x+x)+(x) = 0+(x) = (x).

Finally, given that the additive inverse of x +y is unique, the last claim in (1) follows from the following argument:

(x + y) + (x + y) = (x + y) + (y + x) = x + (y + (y + x)) = x + ((y + y) + x) = x + (0 + x) = x + x = 0.

(Quiz. Prove that 1x = x in any eld. Hint. There is something to be proved here!)

Exercise 25 (Rules of Exponentiation) Let (X , +, ) be a eld. For any x X , we dene x0 := 1, and for any positive integer k, we let xk := xk1x and xk := (xk)1 . For any integers i and j, prove that xix j = xi+j and (xi) j = xij for any x X , and xi/x j = xij and (y/x)i = yi/xi for any x X \{0}.

Although a eld provides a rich environment for doing arithmetic, it lacks structure for ordering things. We introduce such a structure next.

Denition The list (X , +, , ) is called an ordered eld if (X , +, ) is a eld, and if is a partial order on X that is compatible with the operations + and in the sense that x y implies x +z y +z for any x, y, z X , and xz yz for any x, y, z X with z 0. We note that the expressions x y and y x are identical. The same goes also for the expressions x > y and


y < x. 18 We also adopt the following notation:

X+ := {x X : x 0} and X++ := {x X : x > 0}, and

X := {x X : x 0} and X := {x X : x < 0}.

An ordered eld is a rich algebraic system within which many algebraic properties of real numbers can be established. This is of course not the place to get into a thorough algebraic analysis, but we should consider at least one example to give you an idea about how this can be done.

Example 7 (The Triangle Inequality) Let (X , +, , ) be an ordered eld. The function || : X X dened by

x, if x 0|x| := x, if x < 0

is called the absolute value function.19 The following is called the triangle inequality:

x + y |x| + y for all x, y X . You have surely seen this inequality in the case of real numbers. The point is that it is valid within any ordered eld, so the only properties responsible for it are the ordered eld axioms.

We divide the argument into ve easy steps. All x and y that appear in these steps are arbitrary elements of X .

(a) |x| x. Proof. If x 0, then |x| = x by denition. If 0 > x, on the other hand, we have

|x| = x = 0 + x x + x = 0 x. (b) x 0 implies x 0, and x 0 implies x 0. Proof. If x 0,

then

0 = x + x 0 + x = x. 18 Naturally, x > y means that x and y are distinct members of X with x y. That is, > is the asymmetric part of . 19 We owe the notation |x| to Karl Weierstrass. Before Weierstrasss famous 1858 lectures, there was apparently no unity on denoting the absolute value function. For instance, Bernhard Bolzano would write x!

2 Real Numbers | 37

The second claim is proved analogously. (c) x |x|. Proof. If x 0, then x 0 x = |x| where the

second inequality follows from (b). If 0 > x, then |x| = (x) = x by (1).

(d) x y implies y x. Proof. Exercise. y(e) x + y |x| + . Proof. Applying (a) twice, |x| + y x + y = y + x y + x = x + y.

Similarly, by using (c) twice,

x + y |x| + y = |x| + y where we used the third claim in (1) to get the nal equality. By (d), therefore, |x| + y (x + y), and we are done.

Exercise 26 Let (X , +, , ) be an ordered eld. Prove: xy = |x| y and x y |x| y for all x, y X .

2.2 Natural Numbers, Integers, and Rationals

As you already know, we denote the set of all natural numbers by N, that is, N := {1, 2, 3, . . .}. Among the properties that this system satises, a particularly interesting one that we wish to mention is the following:

The Principle of Mathematical Induction If S is a subset of N such that 1 S, and i + 1 S whenever i S, then S = N.

This property is actually one of the main axioms that are commonly used to construct the natural numbers.20 It is frequently employed when giving

20 Roughly speaking, the standard construction goes as follows. One postulates that N is a set with a linear order, called the successor relation, which species an immediate successor for each member of N. If i N, then the immediate successor of i is denoted as i + 1. Then, N is the set that is characterized by the Principle of Mathematical Induction and the following three axioms: (i) there is an element 1 in N that is not a successor of any other element in N; (ii) if i N, then i + 1 N; and (iii) if i and j have the same successor, then i = j. Along with the Principle of Mathematical Induction, these properties are known as the Peano axioms (in honor of Giuseppe Peano (18581932), who rst formulated these postulates and laid out an axiomatic foundation for the integers). The binary operations + and are dened via the successor relation, and behave well because of these axioms.


a recursive denition (as in Exercise 25), or when proving innitely many propositions by recursion. Suppose P1, P2, . . . are logical statements. If we can prove that P1 is true, and then show that the validity of Pi+1 would in fact follow from the validity of Pi (i being arbitrarily xed in N), then we may invoke the Principle of Mathematical Induction to conclude that each proposition in the string P1, P2, . . . is true. For instance, suppose we wish to prove that

1 1 1 1 1 + + + + = 2 for each i N. (2)

2 4 2i 2i

Then we rst check if the claim holds for i = 1. Since 1 + 12 = 2 12 , this is indeed the case. On the other hand, if we assume that the claim is true for an arbitrarily xed i N (the induction hypothesis), then we see that the claim is true for i + 1, because

1 1 1 1 1 1 1 1 + + + + = 1 + + + + +

2 4 2i+1 2 4 2i 2i+1 1 1 = 2 + (by the induction hypothesis) 2i 2i+1 1 = 2 .

2i+1

Thus, by the Principle of Mathematical Induction, we conclude that (2) holds. We shall use this principle numerous times throughout the text. Here is another example.

Exercise 27 Let (X , +, , ) be an ordered eld. Use the Principle of Mathematical Induction to prove the following generalization of the triangle inequality: For any m N,

|x1 + + xm| |x1| + + |xm| for all x1, . . . , xm X .

Adjoining to N an element to serve as the additive identity, namely the zero, we obtain the set of all nonnegative integers, which is denoted as Z+. In turn, adjoining to Z+ the set {1, 2, . . .} of all negative integers (whose construction would mimic that of N), we obtain the set Z of all integers. In the process, the binary operations + and are suitably extended from N to Z so that they become binary operations on Z that satisfy all of the eld axioms except the existence of multiplicative inverse elements.

2 Real Numbers | 39

Unfortunately, the nonexistence of multiplicative inverses is a serious problem. For instance, while an equation like 2x = 1 makes sense in Z, it cannot possibly be solved in Z. To be able to solve such linear equations, we need to extend Z to a eld. Doing this (in the minimal way) leads us to the set Q of all rational numbers, which can be thought of as the collection of all fractions mn with m, n Z and n = 0. The operations + and are extended to Q in the natural way (so that, for instance, the additive and multiplicative inverses of mn are mn and mn , respectively, provided that m, n = 0). Moreover, the standard order on Z (which is deduced from the successor relation that leads to the construction of N) is also extended to Q in the straightforward manner.21 The resulting algebraic system, which we denote simply as Q instead of the fastidious (Q, +, , ), is signicantly richer than Z. In particular, the following is true.

Proposition 4 Q is an ordered eld.

Since we did not give a formal construction of Q, we cannot prove this fact here.22 But it is certainly good to know that all algebraic properties of an ordered eld are possessed by Q. For instance, thanks to Proposition 4, Example 7, and Exercise 25, the triangle inequality and the standard rules of exponentiation are valid in Q.

2.3 Real Numbers

Although it is far superior to that of Z, the structure of Q is nevertheless not strong enough to deal with many worldly matters. For instance, if we take a square with sides having length one, and attempt to compute the length

21 [Only for the formalists] These denitions are meaningful only insofar as one knows the operation of division (and we dont, since the binary operation / is not dened on Z). As noted in Section 1.3, the proper approach is to dene Q as the set of equivalence classes [(m, n)] where the equivalence relation is dened on Z (Z\{0}) by (m, n) (k, l) iff ml = nk. The addition and multiplication operations on Q are then dened as [(m, n)] + [(k, l)] = [(ml + nk, nl)] and [(m, n)][(k, l)] = [(mk, nl)]. Finally, the linear order on Q is dened via the ordering of integers as follows: [(m, n)] [(k, l)] iff ml nk. 22 If you followed the previous footnote, you should be able to supply a proof, assuming the usual properties of Z.


r of its diagonal, we would be in trouble if we were to use only the rational numbers. After all, we know from planar geometry (from the Pythagorean Theorem, to be exact) that r must satisfy the equation r2 = 2. The trouble is that no rational number is equal to the task. Suppose that r2 = 2 holds for some r Q. We may then write r = mn for some integers m, n Z with n = 0. Moreover, we can assume that m and n do not have a common factor. (Right?) Then m2 = 2n2, from which we conclude that m2 is an even integer. But this is possible only if m is an even integer itself. (Why?) Hence we may write m = 2k for some k Z. Then we have 2n2 = m2 = 4k2 so that n2 = 2k2, that is, n2 is an even integer. But then n is even, which means 2 is a common factor of both m and n, a contradiction.

This observation is easily generalized:

Exercise 28 Prove: If a is a positive integer such that a = b2 for any b Z, then there is no rational number r such that r2 = a. 23

Here is another way of looking at the problem above. There are certainly two rational numbers p and q such that p2 > 2 > q2, but now we know that there is no r Q with r2 = 2. It is as if there were a hole in the set of rational numbers. Intuitively speaking, then, we wish to complete Q by lling up its holes with new numbers. And, lo and behold, doing this leads us to the set R of real numbers. (Note. Any member of the set R\Q is said to be an irrational number.)

This is not the place to get into the formal details of how such a completion would be carried out, so we will leave things at this fairy tale level. However, we remark that, during this completion, the operations of addition and multiplication are extended to R in such a way as to make it a eld. Similarly, the order is extended from Q to R nicely, so a great many algebraic properties of Q are inherited by R.

23 This fact provides us with lots of real numbers that are not rational, e.g., 2, 3, 5, 6, . . . , etc. There are many other irrational numbers. (Indeed, there is a sense inwhich

there are more of such numbers than of rational numbers.) However, it is often difcult to prove the irrationality of a number. For instance, while the problem of incommensurability of the circumference and the diameter of a circle was studied since the time of Aristotle, it was not until 1766 that a complete proof of the irrationality of was given. Fortunately, elementary proofs of the fact that / Q are since then formulated. If you are curious about this issue, you might want to take a look at Chapter 6 of Aigner and Ziegler (1999), where a brief and self-contained treatment of several such results (e.g., 2 / Q and e / Q) is given.

2 Real Numbers | 41

Proposition 5 R is an ordered eld.

Notation. Given Propositions 4 and 5, it is natural to adopt the notations Q+, Q++, Q, and Q to denote, respectively, the nonnegative, positive, nonpositive, and negative subsets of Q, and similarly for R+, R++, R, and R.

There are, of course, many properties that R satises but Q does not. To make this point clearly, let us restate the order-theoretic properties given in Exercise 15 for the special case of R. A set S R is said to be bounded from above if it has an -upper bound, that is, if there is a real number a such that a s for all s S. In what follows, we shall refer to an -upper bound (or the -maximum, etc.) of a set in R simply as an upper bound (or the maximum, etc.) of that set. Moreover, we will denote the -supremum of a set S R by sup S. That is, s = sup S iff s is an upper bound of S, and a s holds for all upper bounds a of S. (The number sup S is often called the least upper bound of S.) The lower bounds of S and inf S are dened dually. (The number inf S is called the greatest lower bound of S.)

The main difference between Q and R is captured by the following property:

The Completeness Axiom Every nonempty subset S of R that is bounded from above has a supremum

in R. That is, if = S R is bounded from above, then there exists a real number s such that s = sup S.

It is indeed this property that distinguishes R from Q. For instance, S := {q Q : q2 < 2} is obviously a set in Q that is bounded from above. Yet sup S does not exist in Q, as we will prove shortly. But sup S exists in R by the Completeness Axiom (or, as is usually said, by the com pleteness of the reals), and of course, sup S = 2. (This is not entirely trivial; we will prove it shortly.) In an intuitive sense, therefore, R is obtained from Q by lling the holes in Q to obtain an ordered eld that


satises the Completeness Axiom. We thus say that R is a complete ordered eld. 24

In the rest of this section, we explore some important consequences of the completeness of the reals. Let us rst warm up with an elementary exercise that tells us why we did not need to assume anything about the greatest lower bound of a set when stating the Completeness Axiom.

Exercise 29 H Prove: If = S R and there exists an a R with a s for all s S, then inf S R.

Here is a result that shows how powerful the Completeness Axiom really is.

Proposition 6 (a) (The Archimedean Property) For any (a, b) R++ R, there exists

an m N such that b < ma. (b) For any a, b R such that a < b, there exists a q Q such that

a < q < b. 25

Proof

(a) This is an immediate consequence of the completeness of R. Indeed, if the claim was not true, then there would exist a real number a > 0 such that {ma : m N} is bounded from above. But then s = sup{ma : m N} would be a real number, and hence a > 0 would imply that s a is not an upper bound of {ma : m N}, that is, there exists an m N such that s < (m + 1)a, which is not possible in view of the choice of s.

24 Actually, one can say a bit more in this junction. R is not only a complete ordered eld, it is in fact the complete ordered eld. To say this properly, let us agree to call an ordered eld (X , , , ) complete if sup S X for any S 2X \{} that has an -upper bound in X . It turns out that any such ordered eld is equivalent to R up to relabeling. That is, for any complete ordered eld (X , , , ), there exists a bijection f : X R such that f (x y) = f (x) + f (y), f (x y) = f (x)f (y), and x y iff f (x) f (y). (This is the Isomorphism Theorem. McShane and Botts (1959) prove this as Theorem 6.1 (of Chapter 1) in their classic treatment of real analysis (reprinted by Dover in 2005).) 25 We thus say that the rationals are order-dense in the reals.

2 Real Numbers | 43

(b) Take any a, b R with b a > 0. By the Archimedean Property, there exists an m N such that m(b a) > 1, that is, mb > ma + 1. Dene n := min{k Z : k > ma}. 26 Then ma < n 1 + ma < mb (why?), so letting q := mn completes the proof.

Exercise 30H Show that, for any a, b R with a < b, there exists a c R\Q such that a < c < b.

We will make use of Proposition 6.(b) (and hence the Archimedean Property, and hence the Completeness Axiom) on many occasions. Here is a quick illustration. Let S := {q Q : q < 1}. What is sup S? The natural guess is, of course, that it is 1. Let us prove this formally. First of all, note that S is bounded from above (by 1, in particular), so by the Completeness Axiom, we know that sup S is a real number. Thus, if 1 = sup S, then by denition of sup S, we must have 1 > sup S. But then by Proposition 6.(b), there exists a q Q such that 1 > q > sup S. Yet the latter inequality is impossible, since q S and sup S is an upper bound of S. Hence, 1 = sup S.

One can similarly compute the sup and inf of other sets, although the calculations are bound to be a bit tedious at this primitive stage of the development. For instance, let us show that

sup{q Q : q2 < 2} = 2.

That is, where S := {q Q : q2 < 2}, we wish to show that sup S is a real number the square of which equals 2. Notice rst that S is a nonempty set that is bounded from above, so the Completeness Axiom ensures that s := sup S is real number. Suppose we have s2 > 2. Then s2 2 > 0, so by the Archimedean Property there exists an m N such that m(s2 2) > 2s. Then 2

s 1 = s2 2s + m

12 > s

2 (s2 2) = 2,

m m

which means that s m 1

2 > q2 for all q S. But then s m 1 is an upper

bound for S, contradicting that s is the smallest upper bound for S. It follows that we have s2 2. Good, let us now look at what happens if we have s2 < 2.

26 By the Archimedean Property, there must exist a k N such that k > ma, so n is well-dened.


In that case we use again the Archimedean Property to nd an m N such that m(2 s2) > 4s and m > 21 s . Then 2

s + 1 = s2 + 2s + 1 < s2 + 2s + 2s < s2 + (2 s2) = 2. m m m2 m m

But, by Proposition 6.(b), there exists a q Q with s < q < s + m 1 . It follows that s < q S, which is impossible, since s is an upper bound for S. Conclusion: s2 = 2. Put differently, the equation x2 = 2 has a solution in R, thanks to the Completeness Axiom, while it does not have a solution in Q.

Exercise 31 Let S be a nonempty subset of R that is bounded from above. Show that s = sup S iff both of the following two conditions hold: (i) s s for all s S; (ii) for any > 0, there exists an s S such that s > s . Exercise 32 Let A and B be twononempty subsets of R that arebounded from above. Show that A B implies sup A sup B, and that

sup{a + b : (a, b) A B} = sup A + sup B. Moreover, if c a for all a A, then c sup A. Exercise 33 Let S R be a nonempty set that is bounded from below. Prove that inf S = sup(S), where S := {s R : s S}. Use this result to state and prove the versions of the results reported in Exercises 31 and 32 for nonempty subsets of R that are bounded from below.

2.4 Intervals and R

For any real numbers a and b with a < b, the open interval (a, b) is dened as (a, b) := {t R : a < t < b}, and the